The ring $C^\infty(M)$ of smooth functions on a smooth manifold $M$ is a topological ring with respect to the Whitney topology and the usual ring operations. Is it possible to describe, maybe under some conditions on $M$, the ideals and the closed ideals of $C^\infty(M)$?

unknowngoogle: now that it has been edited to include some precise definitions, the question looks much better. The original version was in itself inoffensive, but the lack of detail was not encouraging
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Yemon ChoiOct 5 '11 at 20:50

2 Answers
2

Let $M$ be an $n$-dimensional manifold.
for each point $p \in M$ and each natural $k$ we define $N(k)$ to be the number of (up to $n$) tuples $m$ such that $|m| \leq k$. Define the map $J_p^k: C^\infty(M) \rightarrow \mathbb{R}^{N(k)}$ by assigning to $f$ the $m$-jets of $f$ at $p$ up to $|m|=k$.

If $I$ is an ideal of $C^\infty(M)$ then its closure is the ideal of functions $f$ such that for each $p$ in $M$ and $k \geq 0$ then $J^k_p f \in J^k_p(I)$.

So in some sense the closed ideals are like $I_\infty$ in Neil's answer.

I was about to upvote this question when I saw the edit history.. I guess I should upvote algori :) EDIT: well, couldn't resist, it's the question we vote, not who asks it.
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Reimundo HeluaniOct 6 '11 at 1:21

For each $n\geq 0$ we have a closed ideal
$$ I_n=\{f: f^{(i)}(0)=0 \text{ for } 0\leq i < n\} $$
We can write $I_\infty$ for the intersection of these, which is again closed. We can also put
$$ J = \{ f : f(x)=0 \text{ for all } x \leq 0\} $$
and note that this is closed and contained in $I_\infty$.

Next, for $n,a>0$ with $n\in\mathbb{Z}$ we can let $K_{n,a}$ be the principal ideal generated by the function $\exp(-a/x^{2n})$. These are all different and contained in $I_\infty$. I am not sure whether they are closed.

For another kind of example, let $\mathcal{U}$ be a free ultrafilter on $\mathbb{R}$ and put
$$ L = \{f : f^{-1}\{0\} \in \mathcal{U} \}. $$
This is a non-closed maximal ideal.

UPDATE:

Now let $A$ be an arbitrary closed ideal in $C^\infty(\mathbb{R})$. Put
$$ X_n = \{ x\in\mathbb{R} : f^{(i)}(x)=0 \text{ for all } i \leq n \text{ and } f\in A\}. $$
Specialising Reimundo's answer to the case $M=\mathbb{R}$, we see that
$$ A = \{ f : f^{(i)}=0 \text{ on } X_n \text{ for all } i\leq n \}. $$
The sets $X_n$ are closed, with $X_n\supseteq X_{n+1}$. Moreover, if $x$ is a non-isolated point of $X_n$ (so it is in the closure of $X_n\setminus\{x\}$) then it is easy to see that $x\in X_{n+1}$. I would guess that the closed ideals biject with chains of subsets with these properties, but I have not tried to prove that.